Countable partitions of Cantor space mod meager Let $I$ be an index set. Given $A\subseteq I\times 2^\omega$ and $i\in I$, set $(A)_i = \{x\in 2^\omega: (i, x)\in A\}$. 
Now let $I\times 2^\omega = \bigcup_{n<\omega} A_n$. How large must $I$ be to ensure that there are indices $i\neq j\in I$ and $n< \omega$ with $(A_n)_i \cap (A_n)_j$ non-meager? In particular, is it consistent that there is a partition of $\omega_1\times 2^w$ into countably many pieces not satisfying the above?
 A: The Borel case. I claim that every uncountable index set $I$ has your property, if the sets $A^i_n$ appearing in the covers are Borel. It follows that the answer to your follow-up particular question about $\omega_1$ is negative for the Borel case. 
Suppose that $I$ is uncountable. Let me set things up a little differently than you did. What you
really have is $I$ many countable coverings of the reals. That
is, for each $i\in I$, we have a countable covering of the
reals $\bigcup_n A^i_n$. 
(So $\bigcup_i (\{i\}\times
A^i_n)$ is what you call $A_n$, but I got confused thinking about it that way.)
We know by the Baire category theorem that for any particular
$i$, there is some non-meager $A^i_{n_i}$ for some $n_i$. Consider the map $i\mapsto
n_i$. Since $I$ is
uncountable, this map cannot be injective, and in fact for
some $n$, it must be that uncountably many $i$ all select the same $n$. So we've got uncountably many $i$ for which $A^i_n$ is non-meager. Since in the Borel sets the meager ideal has the countable chain condition, it cannot be that these sets form an antichain modulo the meager ideal. So there must be $i\neq j$ with $A^i_n\cap A^j_n$ non-meager, and in fact uncountably many like that.
General case. In the general case, the argument shows that any set $I$ whose size is at least the chain condition of the meager ideal will have your property. Basically, we argue just as above. If $\delta$ is the chain condition of the meager ideal, and we have $A^i_n$ with $\bigcup_n A^i_n$ covering the reals, then we get $n_i$ constant for $\delta$ many $i$, contradicting the chain condition, and so $A^i_n\cap A^j_n$ is non-meager for some $i\neq j$. 
Unfortunately, I don't know much about how big the chain condition of the meager ideal can be. There are some results mentioned in Foreman's handbook article.  
A: Let $\kappa$ be least such that whenever $\{A^i_n : n < \omega\}$, for $i < \kappa$ are coverings of $2^{\omega}$, there are $i < j < \kappa$ and $n < \omega$ such that $A^i_n \cap A^j_n$ is non meager.
Claim: CH implies $\kappa > \omega_1$.
Proof: Using an Ulam matrix.
Claim: Assume MA plus not CH. Then $\kappa = \omega_1$.
Proof: Let $\{A^i_n: n < \omega\}$ be coverings of $2^{\omega}$ for $i < \omega_1$. Towards a contradiction, assume that for every $i < j < \omega_1$ and $n < \omega$, $A^i_n \cap A^j_n$ is meager. Put $B^i_n = A^i_n \setminus \bigcup \{A^j_n : j < i\}$ and $B^i = \bigcup \{B^i_n : n < \omega\}$. Then each $B^i$ is comeager and so is $B = \bigcap \{B^i : i < \omega_1\}$ (by MA plus not CH). But now if $x \in B$, then $(\forall i < \omega_1)(\exists n < \omega)(x \in B^i_n)$. Choose $i < j < \omega_1$ and $n < \omega$ such that $x \in B^i_n \cap B^j_n$: Contradiction.
So the lower bound $\omega_1 \leq \kappa$ cannot be improved in ZFC. I am not sure about the optimal upper bound for $\kappa$.
